3.212 \(\int \frac{\csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=82 \[ \frac{2 \cot (c+d x)}{a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{d (a \sin (c+d x)+a)} \]

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(2*a*d) + (2*Cot[c + d*x])/(a*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(2*a*d) + (Cot[c +
d*x]*Csc[c + d*x])/(d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.0895943, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2768, 2748, 3768, 3770, 3767, 8} \[ \frac{2 \cot (c+d x)}{a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(2*a*d) + (2*Cot[c + d*x])/(a*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(2*a*d) + (Cot[c +
d*x]*Csc[c + d*x])/(d*(a + a*Sin[c + d*x]))

Rule 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}-\frac{\int \csc ^3(c+d x) (-3 a+2 a \sin (c+d x)) \, dx}{a^2}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}-\frac{2 \int \csc ^2(c+d x) \, dx}{a}+\frac{3 \int \csc ^3(c+d x) \, dx}{a}\\ &=-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}+\frac{3 \int \csc (c+d x) \, dx}{2 a}+\frac{2 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a d}\\ &=-\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{2 \cot (c+d x)}{a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.511085, size = 85, normalized size = 1.04 \[ -\frac{4 \tan (c+d x)-4 \csc (2 (c+d x))-3 \sec (c+d x)+\csc ^2(c+d x) \sec (c+d x)+3 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \tanh ^{-1}\left (\sqrt{\cos ^2(c+d x)}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

-(-4*Csc[2*(c + d*x)] - 3*Sec[c + d*x] + 3*ArcTanh[Sqrt[Cos[c + d*x]^2]]*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x] + C
sc[c + d*x]^2*Sec[c + d*x] + 4*Tan[c + d*x])/(2*a*d)

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Maple [A]  time = 0.05, size = 115, normalized size = 1.4 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{1}{da \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

1/8/a/d*tan(1/2*d*x+1/2*c)^2-1/2/a/d*tan(1/2*d*x+1/2*c)+2/a/d/(tan(1/2*d*x+1/2*c)+1)-1/8/a/d/tan(1/2*d*x+1/2*c
)^2+1/2/a/d/tan(1/2*d*x+1/2*c)+3/2/a/d*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 0.985118, size = 212, normalized size = 2.59 \begin{align*} -\frac{\frac{\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a} - \frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}{\frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/8*((4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a - (3*sin(d*x + c)/(cos(d*x +
c) + 1) + 20*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3) - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [B]  time = 1.81147, size = 636, normalized size = 7.76 \begin{align*} \frac{8 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right ) - 4}{4 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d +{\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(8*cos(d*x + c)^3 + 6*cos(d*x + c)^2 - 3*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x +
c) - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1
)*sin(d*x + c) - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 2*(4*cos(d*x + c)^2 + cos(d*x + c) - 2)*sin(
d*x + c) - 6*cos(d*x + c) - 4)/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d + (a*d*cos(d*
x + c)^2 - a*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\csc ^{3}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Integral(csc(c + d*x)**3/(sin(c + d*x) + 1), x)/a

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Giac [A]  time = 1.13823, size = 151, normalized size = 1.84 \begin{align*} \frac{\frac{12 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} + \frac{16}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}} - \frac{18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(12*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c))/a^2 + 16/(a*(
tan(1/2*d*x + 1/2*c) + 1)) - (18*tan(1/2*d*x + 1/2*c)^2 - 4*tan(1/2*d*x + 1/2*c) + 1)/(a*tan(1/2*d*x + 1/2*c)^
2))/d